There are several well-known mathematical statements that are 'obvious' but false (such as the negation of the Banach--Tarski theorem). There are plenty more that are 'obvious' and true. One would naturally expect a statement in the latter category to be easy to prove -- and they usually are. I'm interested in examples of theorems that are 'obvious', and known to be true, but that lack (or appear to lack) easy proofs.

Of course, 'obvious' and 'easy' are fuzzy terms, and context-dependent. The Jordan curve theorem illustrates what I mean (and motivates this question). It seems 'obvious', as soon as one understands the definition of continuity, that it should hold; it does in fact hold; but all the known proofs are surprisingly difficult.

A former colleague of mine used to say (to students), "A theorem is obvious if a proof instantly springs to mind," a maxim I like a lot. I think what you are talking about is theorems where a plausible argument instantly springs to mind but falls short of being a proof.
–
gowersJan 9 '11 at 15:12

5

I am tempted to vote to close as subjective and argumentative given the comments on the existing answers. Can we narrow the definition of 'obvious' being used? Something like gowers' definition is good, but depends a lot on one's training. Perhaps something like "if you asked an undergraduate if it were true, they'd bet yes."
–
Qiaochu YuanJan 9 '11 at 16:46

19

I disagree that the Jordan curve theorem is "obvious" but admits a surprisingly difficult proof. The proof for curves with reasonable regularity is not difficult, while the truth of the theorem for wild curves is not so obvious, I think. (At least, I think it is reasonable to argue that most people's sense of this being intuitively clear comes from imagining a rather regular curve in the plane, not a wild one.)
–
EmertonJan 10 '11 at 16:46

52 Answers
52

Any reduced diagram of an alternating link has the fewest possible crossings.

This 19th century conjecture is difficult to prove, with the proof coming only in 1987 by Kauffman, Murasugi, and Thistlethwaite, using the Jones Polynomial. The discovery of this proof was a huge coup for quantum topology; a quantum invariant was used to prove a difficult classical open problem.
While this is certainly hard to prove, I don't think it's unexpectedly hard to prove. Knot diagrams modulo Reidemeister moves form a rather complicated algebraic structure; and there's no reason to expect that any statement about knot diagrams should be easy to prove.

But I consider this a trick answer because the real difficulty is turning this into a completely rigorous statement. Understanding the rigorous definition of connectedness (and understanding the point of making definitions like this) can be a substantial hurdle, but once this hurdle is crossed, the proof is not difficult.

It is obvious that there is a unique point in any given affine plane in a finite-dimensional euclidean vector space which is closest to the origin.

Therefore it would seem similarly obvious that every de Rham cohomology class on a compact oriented riemannian manifold should have a unique representative with minimal $L^2$ norm: namely, its harmonic representative.

Yet it takes some effort (elliptic regularity,...) to prove that the harmonic representative does in fact exist, i.e., that it is a smooth differential form.

Dynamic programing principle (DPP) is one of the 'obvious' and also intuitive one, in the control problem. Many papers proves its validity in various setup, and all proofs are very complicated. But, there is rarely a counter example of DPP. I wonder, if there is general framework on it. See, Dynamic programming principle (DPP)

I must be completely missing the point here. Why is this hard? Can't you just show that there is a small cube inside $S$?
–
Deane YangFeb 28 '11 at 2:14

4

I suppose that this is trivial if one constructs Lebesgue measure by first assigning measure to rectangles and in the end declares a set to be of measure zero if...well, if it has Lebesgue measure zero. If one merely has the classical definition of a zero measure set (a set such that $\forall \varepsilon > 0$ can be covered by countably many rectangles so that the sum of their volumes is at most $\varepsilon$), this is nontrivial. Actually Gowers already mentioned an equivalent result, in which the difficulty is more clear.
–
MarkFeb 28 '11 at 3:34

2

Does this mean that it is hard to show that a rectangle does not have measure zero?
–
Deane YangFeb 28 '11 at 15:54

An asymptotically flat initial data set for the Einstein equations that is sufficiently "close" to the initial data for Minkowski spacetime generates a solution to the Einstein equations that approaches Minkowski spacetime asymptotically. (try saying that fast 3 times)

It is "obvious" because of our physical experience and intuition with gravity, and it is hard to prove because Einstein's equations are quite subtle and complicated.

There are other theorems in mathematical relativity that fall into this category, but this one is especially striking since it is particularly difficult to prove, while it "feels" blatantly true.

Todd, I think something closely resembling David's statement might count as hard: the Mazur-Ulam Theorem, that every bijective isometry between real normed spaces is affine. An indication that it's not obvious is that it's false without the word "bijective": e.g. you can cook up a non-affine isometry from R to (R^2 with the infinity-norm). Reference: helsinki.fi/~jvaisala/mazurulam.pdf . I don't know whether it's appreciably easier if you stick to two dimensions.
–
Tom LeinsterJan 13 '11 at 18:21

In the same genre, if not the same type: The Fundamental Theorem of Algebra. Easily understood by high schoolers, plausible, beautifully simple to state. As far as I know, there are no nice proofs understandable by a good (not brilliant) high school student.

I do not know why this is plausible. What would be an "obvious" reason to expect that a degree 6 polynomial with real coefficients has a complex root?
–
Andres CaicedoJan 10 '11 at 23:30

4

Here is a proof whose main idea is understandable by many high school students. The winding number of the image of the circle of radius $r$ changes from $0$ at $r=0$ to the degree of the polynomial for $r$ large, and it can only change when there is a $0$ of the polynomial.
–
Douglas ZareJan 11 '11 at 1:11

4

If I'm following this correctly, some comments say the example is unsuitable because proving the theorem is actually easy, while the oldest comment says it's unsuitable because it's not obvious. What a mess! Keeping in mind how long it took between the result being conjectured and the first actual proof, I think we are too far removed in time from the result to truly appreciate it from a historical perspective, and the FTA is too fundamentally rooted in students educations to imagine how hard it would be for someone who was a blank slate.
–
Thierry ZellJan 11 '11 at 14:43

1

@Thierry: agreed. Anything Euler attempted unsuccessfully to prove can't be all that easy. I think your comments apply to comments on several other answers as well in relation to how rooted algebraic topology is in many mathematicians' educations these days.
–
Qiaochu YuanJan 12 '11 at 14:34

On an elementary level, the intermediate value theorem is surprisingly deep.

On a less elementary level, the prime number theorem is "obvious" from $\sum_{p\leq x}1/p\sim
\log\log x$ (that was noticed by Euler) and Dirichlet's theorem on primes in arithmetic progressions is "obvious" if you use the sieve of Erathostenes.

P is not equal to NP. This is "obvious", is a straightforward arithmetic proposition doesn't involve any fancy set theory or spacefilling curves, and yet it's so hard that there have whole workshops ("Barriers") and important papers (BGS, natural proofs etc.) devoted to the question of what makes it so hard. Scott Aaronson describes "a would-be P≠NP prover who hasn’t yet grasped the sheer number of mangled skeletons and severed heads that line his path." P≠PSPACE is even more obvious and yet there is a comparable lack of progress.

Very good candidates for this question are theorems that amount to saying that some sequence behaves randomly in a way. Both the fact that the statements are obvious and the fact that they are usually hard to prove, are explained by the fact that there is just no reason for the sequence to nót behave randomly. The primes are of course notorious for this. Easy example that springs to mind: there are about as many primes whose last (decimal) digit is $1$ as there are primes whose last digit is $3$.

Here is an example proof of which is not conceptually difficult, but tedious to write down in full detail: There exists a universal Turing machine.

The reason I find this obvious is that, after learning the definition of what a TM is, a moment's thought will suggest that we should be able to encode information of any TM to a natural number since it is a finitary object. Then, it is conceivable that there is an algorithm which decodes any natural number and carry out the instructions.

Of course, constructing the universal machine and proving that it can actually simulate all the other machines involve some unpleasant details.

I think Godel's completeness theorem is very intuitive. For example, can you imagine a first order theorem that would be true for all groups, that you wouldn't be able to prove (by Godel's definition of `prove'). Of course not! But the proof of the completeness theorem is hard.

I agree that the completeness theorem sounds very intuitive, but I think this is misleading. It takes some serious thought to convince yourself that a particular definition of formal proofs captures mathematical practice (even if you believe intuitively that some definition should work, it's much less obvious that a given deductive system really is complete). Furthermore, I'd bet that many mathematicians would find it equally intuitive that there should be a complete proof system for second-order logic, and of course incompleteness tells us there isn't. So completeness is pretty subtle...
–
Henry CohnSep 8 '12 at 18:17

I'm not sure if it is considered obvious, but I think the Collatz conjecture is also a good example (not just hard to prove, but still unproven!). I've always found it very frustrating that such a simple conjecture does not lend itself to a simple proof.

'Obvious' here sort of means: Try a lot of values, and you'll get a feeling THAT it indeed holds, but also WHY it holds.

No one has mentioned it because it isn't obvious.
–
Johannes HahnFeb 28 '11 at 20:05

3

It appears to be obvious. Combine the obvious fact that more than 3 is necessary with the fact you can't construct a 5 region map that requires 5 colors (follows directly from the fact that the complete graph of 5 node is not plainer) leads most people who look at it casually to mistakenly conclude that 4 is sufficient.
–
Mark BiggarFeb 28 '11 at 21:42

Maybe on the boundary of what's allowed, but I would say most basic geometric things like Pythagoras' Theorem, trigonometry with sine/cosine, the area of a circle, etc. etc. etc.; here of course the difficulty is in defining what we mean by length, area, angle, etc. - in which case some of these become axiomatic, but then the difficulty is shifted onto proving that things do work correctly.

I think this answers your question in a perverse way: All statements in the theory of Natural Numbers provable from the ZFC axioms of set theory. They are obviously true by definition.

EDIT: Looking at this objectively, it probably sounds like I'm saying if a statement is true, then it's obviously true. However, that was not my intent, and I apologize for what may have sounded like a thoughtless response. This is how I see it:

All statements expressible in the language of arithmetic can be represented by formulas in the language of set theory that are only $\Delta_1$ in the Levy hierarchy. In particular, all transitive models will agree on whether they are true. If we further restrict ourselves to only consider the true statements in $\mathbb{N}$ that are ZFC theorems, then all ZFC models will agree that these statements must be true so they are about as obvious to ZFC models as possible. Now if you are an oracle having knowledge of all such true statements, then you will probably develop an intuition that makes them all seem "intuitively obvious." This reflects the answers suggesting that a theorem is obvious after you prove it.

To add one more related point here, when addressing G$\ddot{\textrm{o}}$del's Incompleteness Theorem, one can naively ask about completing PA in the "obvious" way, i.e., by extending it to be the theory consisting of all true statements in $\mathbb{N}$. But of course such a completion is not computable.

It was already obvious to Fermat! ;-) There are different ways to interpret the word "obvious". Not so many people doubt it's true. And just a few understand its proof.
–
Wadim ZudilinJan 9 '11 at 13:48

11

Wadim, sorry for the downvote; but I think you're misunderstanding the English language here. Although the meaning of the statement is obvious, its truth is definitely not. Suppose you'd never heard of it before, and someone asked you to guess the truth or falsity of the statement, in less than an hour, say for a bet of 100 dollars or similar. I cannot believe that any human being could possibly guess "true" and have enough "reasonable" confidence to want to make the bet.
–
Zen HarperJan 9 '11 at 14:06

4

Zen, I guess you are not the one who tells me who could tell me that the Jordan curve theorem is obvious (how many students in your class would agree?). The word "obvious" is too subjective. The things which are obvious to you may be not obvious to others. Therefore, we have a perfect voting system. Everybody feel free to vote down! I have to go to bed to get rid of this nightmare. English language lessons are too much for Sunday. G'night.
–
Wadim ZudilinJan 9 '11 at 14:16

5

Wadim, for what it's worth I also disagree that the Jordan curve theorem is obvious. It's certainly not any more obvious than the nonexistence of space-filling curves.
–
Qiaochu YuanJan 9 '11 at 15:04